(d) (ex + e−x)dy

dx= y2

Section 2.2 Copyright©Arunabha Biswas 61

Section 2.2 Copyright©Arunabha Biswas 62

(d)dy

dx= 2x sin2 y, y(0) =

π

4

Section 2.2 Copyright©Arunabha Biswas 63

Section 2.2 Copyright©Arunabha Biswas 64

(d)dx

dt= 4(x2 + 1), x

(π4

)= 1

Section 2.2 Copyright©Arunabha Biswas 65

Section 2.2 Copyright©Arunabha Biswas 66

Section 2.3 - Linear Equations

Section 2.3 Copyright©Arunabha Biswas 67

Definition (First Order Linear Equation)

A first order ODE of the form

a1(x)dy

dx+ a0(x)y = g(x)

is said to be a first order linear equation in the dependentvariable y.

This ODE can be solved by Integrating Factor Method:

(i) Rewrite the ODE as

dy

dx+ P (x)y = Q(x)

where P (x) =a0(x)

a1(x)and Q(x) =

g(x)

a1(x).

Section 2.3 Copyright©Arunabha Biswas 68

(ii) Multiply both sides of the ODE by the “integrating

factor” or “IF” which is e∫P (x) d(x). So you get

e∫P (x) d(x) dy

dx+ e

∫P (x) d(x)P (x)y = e

∫P (x) d(x)Q(x)

(iii) ⇒ ddx

(ye

∫P (x) dx

)= e

∫P (x) d(x)Q(x)

⇒ d(ye

∫P (x) dx

)= e

∫P (x) d(x)Q(x) dx

⇒∫

d(ye

∫P (x) dx

)=

∫e∫P (x) d(x)Q(x) dx

⇒ ye∫P (x) dx =

∫e∫P (x) d(x)Q(x) dx+ C

Section 2.3 Copyright©Arunabha Biswas 69

Examples 1: Find the general solution of the ODE:

xdy

dx+ 2y = 10x2

Solution:

Section 2.3 Copyright©Arunabha Biswas 70

Section 2.3 Copyright©Arunabha Biswas 71

Examples 3: Solve the IVP:

dy

dx+ y = e−x y(0) = 1

Solution:

Section 2.3 Copyright©Arunabha Biswas 72

Section 2.3 Copyright©Arunabha Biswas 73